The rate constant of a first-order reaction i | vedclass

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Question

(C) For a first-order reaction,the integrated rate equation is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $k = 10^{-3} \ min^{-1}

Vedclass · Accepted Answer

$18$

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(C) For a first-order reaction,the integrated rate equation is given by: $k = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$ Given: $k = 10^{-3} \ min^{-1}$,$t = 200 \ min$. Substituting the values: $10^{-3} = \frac{2.303}{200} \log \frac{[A]_0}{[A]_t}$ $\log \frac{[A]_0}{[A]_t} = \frac{10^{-3} 	imes 200}{2.303} = \frac{0.2}{2.303} \approx 0.0868$ Taking antilog on both sides: $\frac{[A]_0}{[A]_t} = 10^{0.0868} \approx 1.221$ So,$[A]_t = \frac{[A]_0}{1.221} \approx 0.819 [A]_0$ The fraction of reactant reacted is $1 - \frac{[A]_t}{[A]_0} = 1 - 0.819 = 0.181$ Percentage converted = $0.181 	imes 100 = 18.1 \% \approx 18 \%$.

The rate constant of a first-order reaction is $10^{-3} \ min^{-1}$ and the initial concentration of the reactant is $0.2 \ mol \ dm^{-3}$. What percentage of the reactant will be converted into product in $200 \ min$?

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